What is the ampere load of a 3-hp single-phase motor operating at 240 volts with an efficiency rating of 83%?

To determine the ampere load of a 3-horsepower (hp) single-phase motor operating at 240 volts with an efficiency rating of 83%, we start by converting horsepower to watts, since electrical power is typically measured in watts.

One horsepower is equivalent to approximately 746 watts. Therefore, a 3 hp motor will have a power output of:

Power (watts) = 3 hp × 746 watts/hp = 2238 watts.

Next, we need to account for the motor's efficiency. The efficiency indicates how much of the input electrical power is converted into useful mechanical power. In this case, with an efficiency of 83%, the actual input power required by the motor can be calculated as follows:

Input Power = Output Power / Efficiency.

Substituting the values we have:

Input Power = 2238 watts / 0.83 ≈ 2697.59 watts.

Now, to find the current in amperes when the motor is operating at 240 volts, we will use the formula:

Current (amperes) = Power (watts) / Voltage (volts).

Substituting the input power and the voltage:

Current = 2697.59 watts / 240 volts